Question

如图,在正方体ABCD—A₁B₁C₁D₁中,E为AB的中点.

(1)求AD和B₁C所成的角;

(2)证明平面EB₁D⊥平面B₁CD;

(3)求二面角E－B₁C－D的大小(用反三角函数表示)

Answer 1

解法一:(1)解:正方体ABCD—A₁B₁C₁D₁中,AD∥BC,?

∴AD与B₁C所成的角为∠B₁CB或其补角.2分?

∵∠B₁CB=45°,∴AD和B₁C所成的角为45°.                                                           ?

(2)证明:取B₁C的中点F,B₁D的中点G,

连结BF,EG,GF.?

∵CD⊥平面BCC₁B₁,?

∴DC⊥BF.?

又BF⊥B₁C,DC∩B₁C=C,?

∴BF⊥平面B₁CD.                                                                                                         ?

∵GF CD,BECD,??

∴BEGF.∴四边形BFGE是平行四边形.?

∴BF∥GE.∴EG⊥平面B₁CD.                                                                                ?

又EG平面EB₁D,?

∴平面EB₁D⊥平面B₁CD.                                                                                      ?

(3)解:连结EF.

∵CD⊥B₁C,GF∥CD,∴GF⊥B₁C.?

又EG⊥平面B₁CD,EF⊥B₁C,?

∴∠EFG为二面角E-B₁C-D的平面角.                                                                    ?

设正方体的棱长为a,则在△EFG中,?

GF=a,EF=a,

∴cos∠EFG==.                                                                                       

∴二面角E-B₁C-D的大小为arccos.                                                                  ?

解法二:不妨设正方体的棱长为2个长度单位,?

且设=i,=j,=k.?

以i,j,k为坐标向量建立如图所示的空间直角坐标系D—xyz.??

(1)解:∵D(0,0,0),A(2,0,0),C(0,2,0),B₁(2,2,2),?

∴=(2,0,0),=(2,0,2),                                                                            ?

cos〈,〉===.?

∴AD与B₁C所成的角为45°.                                                                                  ?

(2)证明:取B₁D的中点F,连结EF.∵F(1,1,1),E(2,1,0),?

∴=(-1,0,1),=(0,2,0),=0, =0.?                              ?

∴EF⊥CD,EF⊥CB₁.?

∵CD与CB₁相交,∴EF⊥平面B₁CD.                                                                             ?

又EF平面EB₁D,∴平面EB₁D⊥平面B₁CD.                                                               ?

(3)解:设平面B₁CD的法向量M=(1,a,B),?

由?

解得c=0,B=-1,∴M=(1,0,-1).                                                                                   ?

设平面EB₁C的法向量n=(-1,c,D).?

由

解得c=-2,D=1,∴n=(-1,-2,1).                                                                                   ?

∴cos〈m,n〉===-.                                                                   ?

∴二面角EB₁CD的大小为arccos.